我已经阅读了这个网站。从这里我得知,使用cudamallocHost来使用固定的内存比使用cudamalloc会获得更好的性能。然后我使用两个不同的简单程序测试了执行时间:
使用cudaMallocHost:
#include <stdio.h>
#include <cuda.h>
// Kernel that executes on the CUDA device
__global__ void square_array(float *a, int N)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
if (idx<N) a[idx] = a[idx] * a[idx];
}
// main routine that executes on the host
int main(void)
{
clock_t start;
start=clock();/* Line 8 */
clock_t finish;
float *a_h, *a_d; // Pointer to host & device arrays
const int N = 100000; // Number of elements in arrays
size_t size = N * sizeof(float);
cudaMallocHost((void **) &a_h, size);
//a_h = (float *)malloc(size); // Allocate array on host
cudaMalloc((void **) &a_d, size); // Allocate array on device
// Initialize host array and copy it to CUDA device
for (int i=0; i<N; i++) a_h[i] = (float)i;
cudaMemcpy(a_d, a_h, size, cudaMemcpyHostToDevice);
// Do calculation on device:
int block_size = 4;
int n_blocks = N/block_size + (N%block_size == 0 ? 0:1);
square_array <<< n_blocks, block_size >>> (a_d, N);
// Retrieve result from device and store it in host array
cudaMemcpy(a_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost);
// Print results
for (int i=0; i<N; i++) printf("%d %f\n", i, a_h[i]);
// Cleanup
cudaFreeHost(a_h);
cudaFree(a_d);
finish = clock() - start;
double interval = finish / (double)CLOCKS_PER_SEC;
printf("%f seconds elapsed", interval);
}
using malloc
#include <stdio.h>
#include <cuda.h>
// Kernel that executes on the CUDA device
__global__ void square_array(float *a, int N)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
if (idx<N) a[idx] = a[idx] * a[idx];
}
// main routine that executes on the host
int main(void)
{
clock_t start;
start=clock();/* Line 8 */
clock_t finish;
float *a_h, *a_d; // Pointer to host & device arrays
const int N = 100000; // Number of elements in arrays
size_t size = N * sizeof(float);
a_h = (float *)malloc(size); // Allocate array on host
cudaMalloc((void **) &a_d, size); // Allocate array on device
// Initialize host array and copy it to CUDA device
for (int i=0; i<N; i++) a_h[i] = (float)i;
cudaMemcpy(a_d, a_h, size, cudaMemcpyHostToDevice);
// Do calculation on device:
int block_size = 4;
int n_blocks = N/block_size + (N%block_size == 0 ? 0:1);
square_array <<< n_blocks, block_size >>> (a_d, N);
// Retrieve result from device and store it in host array
cudaMemcpy(a_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost);
// Print results
for (int i=0; i<N; i++) printf("%d %f\n", i, a_h[i]);
// Cleanup
free(a_h); cudaFree(a_d);
finish = clock() - start;
double interval = finish / (double)CLOCKS_PER_SEC;
printf("%f seconds elapsed", interval);
}
在两个程序的执行过程中,执行时间几乎相似。实现上有什么问题吗?cudamalloc和cudamallochost的执行有什么确切的区别?
并且随着每次运行,执行时间都会减少。